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Dmitry [639]
3 years ago
12

Select all of the statements that are true.

Physics
1 answer:
natta225 [31]3 years ago
8 0

The correct statements are "Each orbit holds a fixed number of electrons" and "The n=1 orbit can only hold two electrons." According to the Bohr model, the maximum number of electrons that can occupy an orbit is given by 2n^2, where n is the number of the orbit. For instance, when n=1 it means 2n^2= 2(1)^2=2. This particular orbit can only hold up to two electrons. Even though the electrons can gain energy and move to higher orbits or electrons from higher orbits can lose energy and drop to the n=1 level, the energy level would not allow more electrons to enter the orbit once it is full. Again the octet rule, which states that atoms achieve stability by having 8 valence electrons, limits the maximum number of electrons that can be occupied by an orbit. The gain and loss of electrons is done to achieve the noble gas configuration and once that is reached no more electron can be added to an orbit

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Two parallel plates having charges of equal magnitude but opposite sign are separated by 29.0 cm. Each plate has a surface charg
wel

Answer:

(a) E = 3.6 x 10³ N/C = 3.6 KN/C

(b) ΔV = 1044 Volts

(c) K.E = 1.67 x 10⁻¹⁶ J

(d) Vf = 4.47 x 10⁵ m/s

(e) a = 3.45 x 10¹¹ m/s²

(f) F = 5.76 x 10⁻¹⁶ N

(g) E = 3.6 x 10³ N/C = 3.6 KN/C  

(h)  Both values are same in part (h) and (a)

Explanation:

(a)

Electric field between oppositely charged plates is given as follows:

E = σ/ε₀

where,

E = Electric Field Intensity = ?

σ = surface charge density = 32 nC/m² = 3.2 x 10⁻⁸ C/m²

ε₀ = Permittivity of free space = 8.85 x 10⁻¹² C²/N.m²

Therefore,

E = (3.2 x 10⁻⁸ C/m²)/(8.85 x 10⁻¹² C²/N.m²)

<u>E = 3.6 x 10³ N/C = 3.6 KN/C</u>

<u></u>

(b)

E = ΔV/r

ΔV = Er

where,

r = distance between plates = 29 cm = 0.29 m

ΔV = Potential Difference = ?

ΔV = (3.6 x 10³)(0.29)

<u>V = 1044 Volts</u>

<u></u>

(c)

Kinetic Energy of Proton = Work done on Proton

K.E = F r

but,  F = E q

K.E = E q r

where,

q = charge on proton = 1.6 x 10⁻¹⁹ C

Therefore,

K.E = (3600 N/C)(1.6 x 10⁻¹⁹ C)(0.29 m)

<u>K.E = 1.67 x 10⁻¹⁶ J</u>

<u></u>

(d)

K.E = (1/2)m(Vf² - Vi²)

where,

m = mass of proton = 1.67 x 10⁻²⁷ kg

Vf = Final Velocity = ?

Vi = Initial Velocity = 0 m/s

Therefore,

1.67 x 10⁻¹⁶ J = (1/2)(1.67 x 10⁻²⁷ kg)(Vf² - (0 m/s)²]

Vf² = (1.67 x 10⁻¹⁶ J)(2)/(1.67 x 10⁻²⁷ kg)

Vf = √(20 x 10¹⁰ m²/s²)

<u>Vf = 4.47 x 10⁵ m/s</u>

<u></u>

(e)

2as = Vf² - Vi²

2(a)(0.29 m) = (4.47 x 10⁵ m/s)² - (0 m/s)²

a = (20 x 10¹⁰ m²/s²)/0.58 m

<u>a = 3.45 x 10¹¹ m/s²</u>

<u></u>

(f)

F = ma

F = (1.67 x 10⁻²⁷ kg)(3.45 x 10¹¹ m/s²)

<u>F = 5.76 x 10⁻¹⁶ N</u>

<u></u>

(g)

E = F/q

E = (5.76 x 10⁻¹⁶ N)/(1.6 x 10⁻¹⁹ C)

<u>E = 3.6 x 10³ N/C = 3.6 KN/C</u>

<u></u>

(h)

<u>Both values are same in part (h) and (a)</u>

7 0
3 years ago
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